29
$\begingroup$

I decided to follow a recent trend and ask a question about logarithmic integrals :)

Is there a closed form for this integral? $$\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$$

$\endgroup$
8
  • $\begingroup$ See this sci.ccny.cuny.edu/~ksda/PostedPapers/liouv06.pdf and this math.stackexchange.com/questions/155/… $\endgroup$
    – math_man
    Sep 16, 2014 at 18:14
  • $\begingroup$ @math_man There are some general results about non-existence of indefinite integrals in certain terms for some classes of functions, but AFAIK there is no such general theory for definite integrals. $\endgroup$ Sep 16, 2014 at 19:02
  • $\begingroup$ By writing $\log(1+x) = \log(1-x^2) - \log(1-x)$, we can express the integral in terms of derivatives of the Beta function. It is tedious and not very interesting to calculate the result, but it certainly exists in terms of polygamma functions. $\endgroup$
    – user111187
    Sep 16, 2014 at 20:03
  • $\begingroup$ @user111187: I don't think so. $\endgroup$
    – Lucian
    Sep 16, 2014 at 23:36
  • 3
    $\begingroup$ @Lucian Yes, amd also this, this, this, this and this. $\endgroup$
    – Nik Z.
    Sep 19, 2014 at 16:51

2 Answers 2

19
+200
$\begingroup$

This integral is equal to $$ -4\big( \zeta(-3,-1,-1,-1) +\zeta(-3,-1,1,-1) +\zeta(-3,1,-1,1) +\zeta(3,-1,-1,-1) +\zeta(3,-1,1,-1) +\zeta(3,1,-1,1) \big) $$ in terms of the multiple zeta function, which can also be simplified to $$ 2\zeta(-5,-1)-2\zeta(-5,1)+2\zeta(5,-1)+{\textstyle\frac32}\zeta(5,1)+4\zeta(-3,1,1,1), $$ of which only $$ \begin{aligned} \zeta(5,1) &= {\textstyle\frac34}\zeta(6)-{\textstyle\frac12}\zeta(3)^2 \\ \zeta(5,-1) &= {\textstyle\frac{111}{64}} \zeta (6)-{\textstyle\frac{9}{32}} \zeta (3)^2-{\textstyle\frac{31}{16}} \zeta (5) \log (2) \end{aligned} $$ have a known closed form (see also this article about Euler sums, and also Euler Sums and Contour Integral Representations by Philippe Flajolet and Bruno Salvy).


Update (by editor): Based on MZV reduction of weight $6$, expression above is furtherly simplified to: $$-2 \zeta(\bar5,1)+8 \text{Li}_6\left(\frac{1}{2}\right)+4 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+8 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{13 \zeta (3)^2}{16}+\frac{7}{6} \zeta (3) \log ^3(2)-\frac{221 \pi ^6}{30240}+\frac{\log ^6(2)}{9}-\frac{1}{12} \pi ^2 \log ^4(2)$$

$\endgroup$
3
  • $\begingroup$ Could you express it in terms of non-multiple zeta and polylogarithms? $\endgroup$ Sep 21, 2014 at 1:03
  • $\begingroup$ @VladimirReshetnikov No, I tried weight-6 combinations of zeta functions, and logs and polylogs of arguments $\frac{u}{v}$, $1\leq u\leq 4$. But I'm not confident an expression doesn't exist. $\endgroup$
    – Kirill
    Sep 21, 2014 at 1:05
  • $\begingroup$ Perhaps $\zeta(-5,-1)-\zeta(-5,+1)$ could possess a closed form even if neither term separately is known to have one. $\endgroup$
    – Lucian
    Apr 7, 2017 at 21:36
5
$\begingroup$

There is no closed form for this integral as the answer involves $\sum_{n=1}^\infty\frac{H_n}{n^52^n}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$ which have no known closed form and here is how I found them:

Let $I$ denotes our integral $\displaystyle \int_0^1\frac{\ln x\ln^2(1-x)\ln^2(1+x)}{x}\ dx$

Using the algebraic identity

$$12a^2b^2=(a+b)^4+(a-b)^4-2a^4-2b^4$$ and by letting $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write our integral :

$$\small{12I=\underbrace{\int_0^1\frac{\ln x\ln^4(1-x^2)}{x}}_{1-x^2\mapsto x}+\underbrace{\int_0^1\frac{\ln x\ln^4\left(\frac{1-x}{1+x}\right)}{x}}_{\frac{1-x}{1+x}\mapsto x}-2\underbrace{\int_0^1\frac{\ln x\ln^4(1-x)}{x}}_{1-x\mapsto x}\ dx-2\int_0^1\frac{\ln x\ln^4(1+x)}{x}\ dx}$$

$$12I=-\frac74\underbrace{\int_0^1\frac{\ln(1-x)\ln^4x}{1-x}\ dx}_{K}+2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^4x}{1-x^2}\ dx}_{J}-2\underbrace{\int_0^1\frac{\ln x\ln^4(1+x)}{x}\ dx}_{M}$$


$$K=\int_0^1\frac{\ln(1-x)\ln^4x}{1-x}\ dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^4x\ dx\\ =-24\sum_{n=1}^\infty\frac{H_n}{(n+1)^5}=-24\sum_{n=1}^\infty\frac{H_n}{n^5}+24\zeta(6)=\boxed{12\zeta^2(3)-18\zeta(6)}$$


To evaluate $J$ we are going to use the identity

$$\frac{1}{1-x^2}\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)x^{2n-1}$$

$$J=\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^4x}{1-x^2}\ dx=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)\int_0^1x^{2n-1}\ln^4x\ dx\\ \sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)\left(\frac{3}{4n^5}\right)=-\frac{93}{4}\sum_{n=1}^\infty\frac{H_n}{n^5}-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}\\ =\boxed{\frac{93}{8}\zeta^2(3)-\frac{651}{16}\zeta(6)-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}}$$


I managed to simplify $M$ here

$$M=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}$$


Combining the results of $K$, $J$ and $M$ we get

$$I=20\operatorname{Li}_6\left(\frac12\right)+12\ln2\operatorname{Li}_5\left(\frac12\right)+4\ln^22\operatorname{Li}_4\left(\frac12\right)-\frac{549}{32}\zeta(6) -\frac18\ln2\zeta(5)+\frac14\ln^22\zeta(4)\\ +\frac12\ln^32\zeta(3)-\frac13\ln^42\zeta(2)-\frac{29}{16}\zeta^2(3)+2\ln2\zeta(2)\zeta(3)\\ +\frac{17}{180}\ln^62-4\sum_{n=1}^\infty\frac{H_n}{n^52^n}-4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$

and here we see the two sums appeared and because their numerical values (given by wolfram) are different so unfortunately they don't cancel each other out. So the integral $I$ has no closed form.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .